OFFSET
0,2
COMMENTS
x*G(x) = (1 - 2*x - sqrt(1 - 8*x + 4*x^2))/2 is the o.g.f. of A047891.
The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all primes p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for all primes p >= 5 and positive integers n and k. Examples of these supercongruences are given below.
More generally, for each integer m, we conjecture that the sequence {a_m(n) : n >= 0}, defined by setting a_m(n) = the n-th order Taylor polynomial of G(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences.
FORMULA
a(n) = [x^n] ( (1 + x)*G(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*F'(x)/F(x) )/( 1 - x*F(x) ), where F(x) = (1/x)*Revert( x/G(x) ) = = 1 + 3*x + 21*x^2 + 192*x^3 + 2001*x^4 + ... is the o.g.f. of A219535.
Row sums of the Riordan array ( 1 + x*F'(x)/F(x), x*F(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ sqrt(1 + 17/sqrt(33)) * (59 + 11*sqrt(33))^n / (sqrt(3*Pi*n) * 2^(3*n + 3/2)). - Vaclav Kotesovec, May 02 2024
EXAMPLE
n-th order Taylor polynomial of G(x)^n:
n = 0: G(x)^0 = 1 + O(x)
n = 1: G(x)^1 = 1 + 3*x + O(x^2)
n = 2: G(x)^2 = 1 + 6*x + 33*x^2 + O(x^3)
n = 3: G(x)^3 = 1 + 9*x + 63*x^2 + 414*x^3 + O(x^4)
n = 4: G(x)^4 = 1 + 12*x + 102*x^2 + 768*x^3 + 5493*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 3 = 4, a(2) = 1 + 6 + 33 = 40, a(3) = 1 + 9 + 63 + 414 = 487 and a(4) = 1 + 12 + 102 + 768 + 5493 = 6376.
The triangle of coefficients of the n-th order Taylor polynomial of G(x)^n, n >= 0, in descending powers of x begins
row sums
n = 0 | 1 1
n = 1 | 3 1 4
n = 2 | 33 6 1 40
n = 3 | 414 63 9 1 487
n = 4 | 5493 768 102 12 1 6376
...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 157349171819155 - 4 = (3^3)*(13^3)*269*9860941 == 0 (mod 13^3).
a(2*7) - a(2) = 2314894133906086 - 40 = 2*(3^4)*(7^3)*11*12119*312509 == 0 (mod 7^3).
MAPLE
G := x -> (1/2)*(1 - 2*x - sqrt(1 - 8*x + 4*x^2))/x:
H := (x, n) -> series(G(x)^n, x, 41):
seq(add(coeff(H(x, n), x, k), k = 0..n), n = 0..20);
MATHEMATICA
Table[SeriesCoefficient[(2*(1 + x)^2/(1 - x + Sqrt[1 - 6*x - 3*x^2]))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 02 2024 *)
PROG
(PARI) G(x) = (1 - 2*x - sqrt(1 - 8*x + 4*x^2))/(2*x);
a(n) = my(x='x+O('x^(n+2))); subst(Pol(Vec(G(x)^n)), 'x, 1); \\ Michel Marcus, May 07 2024
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Apr 23 2024
STATUS
approved